Question #b226c

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Question #b226c

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Answer 1 · 2017-04-22T12:30:00

$K_{sp} = 1.6 \times 10^{- 9}$ $K_"sp"=1.6xx10^-9$ .............

Explanation:

$5 C_{2} O_{4}^{2 -} + 2 M n O_{4}^{-} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} (g) ↑ ⏐ ⏐ + 8 H_{2} O$ $5C_2O_4^(2-) + 2MnO_4^(-) + 16H^(+) rarr 2Mn^(2+) + 10CO_2(g)uarr + 8H_2O$ .

So, we works out moles of $permanganate$ $"permanganate"$ :

$= 0.001 \cdot m o l \cdot L^{- 1} \times 4 \times 10^{- 3} L = 4 \times 10^{- 6} \cdot m o l$ $=0.001*mol*L^-1xx4xx10^-3L=4xx10^-6*mol$

And thus, by the stoichiometry, there were $\frac{5}{2} \times 4 \times 10^{- 6} \cdot m o l$ $5/2xx4xx10^-6*mol$ $calcium oxalate$ $"calcium oxalate"$ , which were dissolved in $250 \cdot m L$ $250*mL$ water under conditions of saturation, i.e. a concentration of.......

$\frac{\frac{5}{2} \times 4 \times 10^{- 6} \cdot m o l}{0.250 \cdot L} = 4.0 \times 10^{- 5} \cdot m o l \cdot L^{- 1}$ $(5/2xx4xx10^-6*mol)/(0.250*L)=4.0xx10^-5*mol*L^-1$ .

But, by specification, this represented a saturated solution.

And so $K_{sp} = [C a^{2 +}] [C_{2} O_{4}^{2 -}]$ $K_"sp"=[Ca^(2+)][C_2O_4^(2-)]$

$= {(4.0 \times 10^{- 5} \cdot m o l \cdot L^{- 1})}^{2} = 1.6 \times 10^{- 9}$ $=(4.0xx10^-5*mol*L^-1)^2=1.6xx10^-9$

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Question #b226c

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